Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
注意,链表循环并不是尾指针和头指针相同,可能是在中间某一段形成一个环路,所以不能只判断元素和第一个元素是否存在重合
先设置两个指针p_fast和p_slow。从头开始遍历链表,p_fast每次走两个节点,而p_slow每次走一个节点,若存在循环,这两个指针必定重合:
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 bool hasCycle(ListNode *head) { 12 if(head == NULL) return false; 13 14 ListNode *p_fast = head; 15 ListNode *p_slow = head; 16 17 do{ 18 p_slow = p_slow->next; 19 if(p_fast != NULL) 20 p_fast = p_fast->next; 21 if(p_fast != NULL) 22 p_fast = p_fast->next; 23 else 24 return false; 25 }while(p_fast != p_slow); 26 27 return true; 28 29 } 30 };